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Root locus

Given the open loop transfer function GH

1) asymptotic angles θi​=n−m180−360i​

where n: num of poles, m: num of zeros, i from 0 to n−m−1

2) asymptotic centroid σ=n−m∑poles−∑zeros​

3) break points dsd​K=0 or dsd​GH and solve for S

4) departure angle (ϕ):∑ poles angles∣pole​−∑ zeros angles∣pole​+ϕ=180∘

5) arrival angle (ϕ):∑ poles angles∣zero​−∑ zeros angles∣zero​−ϕ=180∘

Ex) Draw root locus plot for GH=s(s2+8s+32)s2−8s+32​

poles: 0,−4±4j zeros: 4±4j

asymptotic angle =3−2180+(360×0)​=180∘

centroid =3−2(−4+4j)+(−4−4j)−((4+4j)+(4−4j))​=−16

break points dsd​K=0K=−GH1​=−s2−8s+32s3+8s2+32s​

(s2−8s+32)(3s2+16s+32)−(s3+8s2+32s)(2s−8)=0

s4−16s3+512s+1024=0⟹s=−2.928 or s=10.928 (rejected)

departure angle for (−4+4j):

(90+135)−(180+135)+ϕ=180∘⟹ϕ=270∘=−90∘

departure angle for (−4−4j):ϕ=−270∘=90∘

arrival angle for (4+4j):

(0+45+45)−(90)−ϕ=180∘⟹ϕ=−180∘=180∘

arrival angle for (4−4j):ϕ=180∘

to findKcritical​:

characteristic equation 1+GH=0

K(s2−8s+32)+s3+8s2+32s=0

s3+(K+8)s2+(−8K+32)s+32K=0

(k+8)(−8k+32)−32k=0⟹kcritical​=2.928

(k+8)s2+32k=0

10.928s2+32×2.928=0⟹s=±j2.928

or by substituting s=jω in the characteristic equation

s3+(K+8)s2+(−8K+32)s+32K=0

−jω3−(K+8)ω2+(−8K+32)jω+32K=0

real = 0

−(K+8)ω2+32K=0⟹ω2=K+832K​

imaginary = 0

−ω3+(−8K+32)ω=0⟹ω2=−8K+32

equate both values

K+832K​=−8K+32

−8K2+32K−64K+256=32K

−8K2−64K+256=0⟹Kcritical​=2.928

substituting k in ω2=−8K+32

ω=√32−8×2.928​=±2.928

Range of K for system stability 0<K<2.928